Series
A series usually defined as the sum of the terms in an infinite sequence. A series is considered convergent if the sequence of partial sums approaches a specific value and divergent if it approaches positive or negative infinity or if it does not approach any value at all. Divergence and convergence tests Geometric series If a series is geometric, or in the form : \sum_{n=0}^\infty ar^n it will be convergent if |r|<1 . The sum will be equal to : \frac{a}{1-r} Comparison test If a series converges, and every term in a second series is smaller than all corresponding terms in the first series, the second series must also converge. The opposite is also true (if the first series diverges and all the terms in the second are larger, the second series diverges). Limit comparison test (proof) If a_n,b_n>0 , \lim_{n\to\infty}\frac{a_n}{b_n} is finite and not equal to 0, then \sum_{n=1}^\infty a_n converges if and only if \sum_{n=1}^\infty b_n converges. Limit test (proof) If the limit of the sequence being summed does not equal 0, the series diverges. : \lim_{n\to\infty} a_n = 0 This is more useful for proving divergence. Root test If the limit L=\lim_{n\to\infty}\sqrtn <1 , the series \sum_{n=1}^\infty a_n is convergent. If it is greater than one, it will diverge, and if it is equal to one, the test is inconclusive and the series could be either divergent or converge. Integral test (proof) If f(n)=a_n and -\infty<\int\limits_a^\infty f(x)dx<\infty , \sum_{n=a}^\infty a_n must converge. It is important to note that : \int\limits_a^\infty f(x)dx\ne\sum_{n=a}^\infty a_n in general,however the integral test does put some bounds on the series if it is convergent. : \int_{a}^{\infty} f(x) dx \leq \sum_{n=a}^\infty a_n \leq f(a) + \int_{a}^{\infty} f(x) dx Ratio test Given the series \sum_{n=k}^\infty a_n , and \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=r #If r<1 the series converges. #If r>1 the series diverges. #If r=1 the the test is inconclusive and the series could be either divergent or converge. This test is also known as d'Alembert's ratio test or as the Cauchy ratio test. Alternating series test An alternating series, or one in the form \sum_{n=1}^\infty(-1)^n a_n or \sum_{n=1}^\infty(-1)^{n-1}a_n will converge if # |a_n|\ge |a_{n+1}| for all n # \lim_{n\to\infty}a_n=0 Interval of convergence If the terms of the sum have a variable in them, the interval over which the sum converges is called the interval of convergence. This interval can be found by taking the ratio test \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=r . The interval of convergence is the interval of the variable in which |r|<1 . For example, : \sum_{n=k}^\infty\frac{1}{nx^n} : : |r|<1 when |x|>1 , so the interval of convergence is x<-1,x>1 . This test does not tell us anything about the endpoints, so they must be tested separately. When x=1 , we get the harmonic series which is divergent. When x=1 we get the alternating harmonic series, which is convergent. This means the interval of convergence is (-\infty,-1]\cap(1,\infty) . Absolute versus conditional convergence A series is said to converge conditionally if the sum converges but the sum of the absolute value of the terms does not, and a series converges absolutely if the absolute value of the terms converges as well. Since conditionally convergent series are most often seen in the form of alternating series, the alternating series test is typically the most useful in finding the type of convergence. Applications One of the most important applications of series are Taylor series, which can be used to approximate transcendental functions and transcendental numbers. Another important series is the Fourier series, which is used to approximate periodic functions. Examples Does the series below converge? : \sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots We can use the integral test here. Since \int\limits_a^\infty f(x)dx=\int\limits_a^\infty\frac{dx}{n}=\infty , this series diverges. This series is known as the harmonic series. ---- Does the series below converge? : \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots Now we can use the alternating series test. Since a_n\ge a_{n+1} for all n , and \lim_{n\to\infty}\frac{1}{n}=0 , this series converges, although not absolutely, since : \sum_{n=1}^\infty\left|\frac{(-1)^{n-1}}{n}\right|=\sum_{n=1}^\infty\frac{1}{n}=\infty See also *Sequence *Summation *Product operator Category:Calculus Category:Series